In a circuit for finding the resistance of a galvanometer by half deflection method, a 6 V battery and a high resistance of 11 k \Omega are used. The figure of merit of the galvanometer is 60 \mu A/ division. In the absence of shunt resistance, the galvanometer produces a deflection of \theta = 9 divisions when current flows in the circuit. The value of the shunt resistance that can cause the deflection of \theta /2, is closest to :

Current Electricity PYQ — JEE Main 2018 · 16 Apr · Shift 1 · Q74 with Solution

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Correct answer: D

Given data

Battery voltage: $E = 6\,\text{V}$
Series high resistance: $R = 11\,\text{k}\Omega = 11000\,\Omega$
Figure of merit of galvanometer: $k = 60\,\mu\text{A/div}$
Initial deflection: $\theta = 9$ divisions

So initial current through galvanometer is

I_g = k\theta = 60\times 9\,\mu\text{A} = 540\,\mu\text{A} = 5.4\times 10^{-4}\,\text{A}

Find galvanometer resistance $G$

Initially, no shunt is connected, so the galvanometer is in series with $R$ .

Hence,

I_g = \frac{E}{R+G}

Therefore,

R+G = \frac{E}{I_g} = \frac{6}{5.4\times 10^{-4}} = 11111.1\,\Omega

So,

G = 11111.1 - 11000 = 111.1\,\Omega

Thus galvanometer resistance is approximately

G \approx 111\,\Omega

Condition for half deflection

When a shunt $S$ is connected across the galvanometer, the deflection becomes $\theta/2$ .

Since deflection is proportional to current through galvanometer,

I_g' = \frac{I_g}{2} = 270\,\mu\text{A}

In half-deflection method, if the series resistance $R$ is very large compared to $G$ , the total current remains nearly unchanged, so the other half of the current goes through the shunt.

Thus current through shunt is also approximately

I_s \approx I_g' = \frac{I_g}{2}

Since galvanometer and shunt are in parallel, potential drop across them is same:

I_g' G = I_s S

With $I_s = I_g'$ , we get

S = G \approx 111\,\Omega

Closest option

S \approx 110\,\Omega

So the correct option is D.

Verification by exact formula

Using exact half-deflection relation:

S = \frac{RG}{R+G}

Substitute $R=11000\,\Omega$ , $G=111.1\,\Omega$ :

S = \frac{11000\times 111.1}{11111.1} \approx 110\,\Omega

This confirms the result.

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View written solutionFree

Correct answer: D

Given data

Battery voltage: $E = 6\,\text{V}$
Series high resistance: $R = 11\,\text{k}\Omega = 11000\,\Omega$
Figure of merit of galvanometer: $k = 60\,\mu\text{A/div}$
Initial deflection: $\theta = 9$ divisions

So initial current through galvanometer is

I_g = k\theta = 60\times 9\,\mu\text{A} = 540\,\mu\text{A} = 5.4\times 10^{-4}\,\text{A}

Find galvanometer resistance $G$

Initially, no shunt is connected, so the galvanometer is in series with $R$ .

Hence,

I_g = \frac{E}{R+G}

Therefore,

R+G = \frac{E}{I_g} = \frac{6}{5.4\times 10^{-4}} = 11111.1\,\Omega

So,

G = 11111.1 - 11000 = 111.1\,\Omega

Thus galvanometer resistance is approximately

G \approx 111\,\Omega

Condition for half deflection

When a shunt $S$ is connected across the galvanometer, the deflection becomes $\theta/2$ .

Since deflection is proportional to current through galvanometer,

I_g' = \frac{I_g}{2} = 270\,\mu\text{A}

In half-deflection method, if the series resistance $R$ is very large compared to $G$ , the total current remains nearly unchanged, so the other half of the current goes through the shunt.

Thus current through shunt is also approximately

I_s \approx I_g' = \frac{I_g}{2}

Since galvanometer and shunt are in parallel, potential drop across them is same:

I_g' G = I_s S

With $I_s = I_g'$ , we get

S = G \approx 111\,\Omega

Closest option

S \approx 110\,\Omega

So the correct option is D.

Verification by exact formula

Using exact half-deflection relation:

S = \frac{RG}{R+G}

Substitute $R=11000\,\Omega$ , $G=111.1\,\Omega$ :

S = \frac{11000\times 111.1}{11111.1} \approx 110\,\Omega

This confirms the result.

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